Problem: $\overline{AC}$ is $16$ units long $\overline{BC}$ is $12$ units long $\overline{AB}$ is $20$ units long What is $\cot(\angle ABC)?$ $A$ $C$ $B$ $16$ $12$ $20$
$\cot(\angle ABC) = \dfrac{1}{\tan(\angle ABC)}$ How can we find $\tan(\angle ABC)$ SOH CAH TOA angent = pposite over djacent Opposite $= \overline{AC} = 16$ Adjacent $= \overline{BC} = 12$ $\tan(\angle ABC) = \dfrac{16}{12}$ $\cot(\angle ABC) = \dfrac{1}{\tan(\angle ABC)} = \dfrac{12}{16}$